Gearbox design

1. Theory and Design of mechanical system Topic: 2 stage cylindrical-bevel speed reducer Group: G2 1. 111910138 Michael Susngi 2. 111910139 Pranav Jadhav 3. 111910140 Rahul Jadhav 4. 111910141 Ravindra Shinde Team members

2. ● Problem Statement “Design a suitable gearbox for box shipping conveyor which used to carry 3 boxes of 50kg each at a speed of 28m/min.”

4. • Specifications : 1. Load acting on belt: 3 x 50kg 2. Speed of conveyor 3. Diameter of roller 4. Weight of conveyor belt 5. Friction coefficient between belt & roller • Assumptions : 1. Material is assumed to be isotropic and homogeneous. 2. All types of vibrations are neglected. 3. Thermal stresses are neglected. 4. Gradual and uniform loading condition. 150kg 28m/min 200mm 50kg 0.67 (Reference : https://www.rubbercal.com/industrial-rubber/heavy-black-conveyor-belt.html & https://www.engineersedge.com/coeffients_of_friction.htm )

5. ● Design Requirements 1. Nominal operating conditions of industrialized locations 2. Input and output shafts STD. size for typical couplings 3. Usually low shock levels, occasional moderate shock 4. Continuous operation 5. Low maintenance 6. Competitive cost

6. • Significance of gear drive: 1. It is a positive drive, and the velocity ratio remains constant. 2. The center distance between the shafts is relatively small, which results in compact construction. 3. It can transmit very large power, which is beyond the range of belt or chain drives. 4. It can transmit motion at very low velocity, which is not possible with the belt drives. 5. The efficiency of gear drives is very high, even up to 99 per cent in case of spur gears. 6. A provision can be made in the gearbox for gear shifting, thus changing the velocity ratio over a wide range.

7. • Types of gear: 1. Spur gear 1. Easy to design and manufacture. 2. Suitable for low torque 3. Make noise at high-speed operation 2. Bevel gear 1. Angular torque transmission. 2. Not desirable for High-speed reduction. 3. Compact design and highly efficient

8. 4. Helical gear 3. Worm and Worm wheel 1. Large speed reduction. 2. Large increase in torque. 3. Self-Locking Mechanism. 4. Less efficient 1. Gradual teeth engagement 2. More durable 3. Less noise 4. Produce large thrust force due to helix angle • Types of gear:

9. • Selection of gearbox 2 stage cylindrical bevel speed reducer. a) In our case motor is in parallel to the conveyor belt. so, we have to transmit power in perpendicular direction. So perfect solution is bevel gear. b) But along with that we have to achieve high reduction ratio of 30. so, we also used spur gears to obtain the desired reduction.

10. • Orthographic views with notations:

11. PART NO. PART NAME TYPE QTY 1 Driving spur gear Custom 1 2 Driven spur pinion Custom 1 3 Driving bevel pinion Custom 1 4 Driven bevel gear Custom 1 5 Input shaft Custom 1 6 Output shaft Custom 1 7 Intermediate shaft Custom 1 8 Input shaft Bearings STD. 2 9 Output shaft Bearings STD. 2 10 Intermediate shaft Bearings STD. 2 11 Upper casing Custom 1 12 Lower casing Custom 1 • Part list : PART NO. PART NAME TYPE QTY 13 Casing Nut STD. 4 14 Casing bolt STD. 4 15 Foundation bolt STD. 4 16 Bearing cap bolt STD. 20 17 Input shaft oil seal STD. 1 18 Output shaft oil seal STD. 1 19 Banjo bolt STD. 1 20 Input shaft key STD. 1 21 Output shaft key STD. 1 22 Intermediate shaft key 1 STD. 1 23 Intermediate shaft key 2 STD. 1 24 Retaining ring STD. 3 25 Bearing cap Custom 5

12. • Selection of motor i) Power required to convey boxes = µN x V = 0.67 x (150+50) x 9.81 x 28 60 = 613.5 watt ii) To encounter different losses in the gearbox and couplings 20% overload factor is considered. Thus, total power required = 613.36 + 613.36 x 20% = 736.03 watt iii) Reasons for using Induction motor 1. Useful in low-speed operation 2. Frequent maintenance not needed as like DC (Due to carbon brushes) 3. Suitable in operations where high starting torque is not required such as in conveyors

13. The standard motor available in induction motor catalogue to obtain required power is of 0.75kW(1HP) (Reference: https://3.imimg.com/data3/QM/MP/MY-3212916/three-phase-induction-motors.pdf, Table no. 3, pg. No. 3) “3-phase Squirel Cage Induction Motor, 50Hz, 0.75kW(1HP)” • Selection of motor Electrical Specification Mechanical Specification Number of poles : 8 Speed : 690 RPM Full load current : 2A Efficiency : 70.0% Operating voltage : 415V ± 10% Output : 0.75 kW Power factor : 0.73 Frame : 100L

14. Name of components Abbreviation used Input shaft P Driving Bevel Pinion G1 Driven Bevel gear G2 Intermediate Shaft Q Driving Spur pinion S1 Driven Spur gear S2 Output shaft R Input Shaft Bearing B1 Output Shaft Bearing B2 Intermediate Shaft Bearing B3 • Gear train Layout with abbreviations used

15. MOTOR ROLLER • Power flow diagram Motor Coupling 1 P G1 G2 Q S1 S2 R ROLLER Coupling 2

16. • Force analysis of whole system (3D)

17. • Gear ratio calculations 1. After selection of motor, Input speed to gearbox = 690 rpm 2. Speed at output of gearbox is calculated as; Velocity of roller = radius of roller x angular velocity of roller V = r x ( 2π𝑁 60 ) ; ∴ 𝑁 = 𝑉 ×60 2𝜋𝑟 = 28×60 60×2𝜋×0.2 = 23.3 𝑟𝑝𝑚 Thus, Output speed of gearbox = 23.3 rpm 3. That is, we have to reduce speed from 690rpm to 23.3rpm in 2 stages 4. After doing several permutations and combinations, we get the suitable configuration of, 5:1 in 1st stage (in bevel pair) ∴ Speed of shaft Q = 690 5 = 138 rpm & 6:1 in 2nd stage (in spur pair) ∴ Speed of shaft R = 138 6 = 23 rpm Shaft Type Speed (RPM) P 690 Q 138 R 23

18. • Bevel pair design i) Part No. is 3,4 and part name is Driving bevel pinion and Driven bevel gear ii) Function: Transmits power in perpendicular direction and also provides speed reduction iii) Free body diagram Material Syt (𝑘𝑔𝑓/𝑚𝑚2 ) Sut (𝑘𝑔𝑓/𝑚𝑚2 ) Pinion 40 Ni 1 Cr 1 Mo 28 88 110 Wheel C 55 Mn75 46 76 iv) For reduction ratio of 5:1 material selected are; (Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.12 ,1.15)

19. • Bevel pair design v) Two orthographic views SEC. F. V. SEC. R. H. V.

20. vi) Design calculations: Let, rpm = 695 , module : 3 , reduction ratio i = 5 , Number of teeth on pinion Zp = 20

23. vii) Verifying Design Against Beam Strength:

24. viii) Verifying Design Against Wear Strength

25. ix) Final dimensions: Notations Meaning Values 𝑚 module 3 𝛼 Pressure angle 20˚ 𝑍𝑝 Num of teeth on pinion 20 𝑍𝐺 Num of teeth on gear 100 𝐷𝑝 p.c.d. of pinion (G1) 60 mm 𝐷𝐺 p.c.d. of gear (G2) 300mm 𝐴𝑜 Cone distance 153 mm 𝑏 Face width 38.25 mm 𝛾 Pitch angle 11.30˚

26. • Spur pair design i) Part No. is 1,2 and part name is Driving spur pinion and Driven spur gear ii) Function: Transmits power in parallel shafts and speed reduction iii) Free body diagram Material Syt (𝑘𝑔𝑓/𝑚𝑚2 ) Sut (𝑘𝑔𝑓/𝑚𝑚2 ) Pinion (S1) 40 Ni 1 Cr 1 Mo 28 88 110 Wheel (S2) C 35 Mn75 46 76 iv) For reduction ratio of 6:1 material selected are; (Refer: PSG data book, Table 5, pg.no. 8.4 & PSG data book, pg.no. 1.9)

27. • Spur pair design v) Two orthographic views

28. vi) Design calculations (Beam & wear Strength)

29. vi) Design calculations (Beam & wear Strength)

30. vi) Final dimensions of Spur gear Notations Meaning Values 𝑚 module 3 𝛼 Pressure angle 20˚ ℎ Tooth depth (both) 6.75 mm 𝑍1 Num of teeth (pinion) 18 𝑍2 Num of teeth (Gear ) 108 𝐷1 P.C.D. of pinion (G1) 54 mm 𝐷2 P.C.D. of gear (G2) 324 mm 𝐷𝑎 Addendum Diameter(Gear) 330 mm 𝑑𝑎 Addendum Diameter(pinion) 60 mm 𝐷𝑏 Base diameter(Gear) 316.5 mm 𝑑𝑏 Base diameter (pinion) 46.5 mm

31. • Design of shafts 1. Intermediate shaft i) Part no. is 7 and part name is intermediate shaft ii) Design of based on ASME code iii) Function: transmission of torque and power

32. iv) The material selected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) v) Properties of material: 1. Sut = 580N/mm2 2. Syt = 330N/mm2 vi) Allowable stress calculation: Ʈmax = 0.30xSyt = 99N/mm2 or Ʈmax = 0.18xSut = 104.4N/mm2 hence Ʈmax = 99N/mm2 … (which is less) Also, shaft is keyed for mounting of gears, ∴ Ʈmax = 0.75x99 = 74.25N/mm2

33. vii) FBD & BMD of shaft in respective plane ∴ Maximum bending moment at S1 = 44384.182 + 126856.842 = 134,397.22 Nmm

34. viii) Thus, Mb = 134397.22 Nmm & Mt = 77887 Nmm ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 134397.22)2+(1.5 × 77887)2 ∴ 𝑑 = 25.19𝑚𝑚 But we haven’t considered the weight of gears while calculating diameter of shaft, hence we consider the shaft of diameter d = 35mm xi) Finally, Shaft dia. for bearing mounting d 35mm Shaft dia. for gear mounting 1.1d 38.5mm Shaft dia. for resting gear on one side 1.2d 42mm

35. xii) Checking stress concentration at critical sections As there are steps/shoulders in shaft ; there has to be a stress concentration. To reduce this stress concentration fillet is provided From BMD it is clear that on spur gear(S1) side bending moments are higher. Hence critical sections X-X’ and Y-Y’ are labelled in adjoining figure We have to find stress(𝜎)at these sections and then comparing with Endurance limit (Se ) If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.1d,1.2d and if σ < Se , then we can say that Design is safe and correct.

36. A) At section X-X’ for Bending and Torsion

37. B) At section Y-Y’ for Bending and Torsion

38. • Selection of bearings 1. Intermediate shaft bearings i) Part No. 3 and part name is Intermediate shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1)

39. The construction, which involves two bearings with their fronts facing each other, is called ‘face-to-face’ or direct mounting. (Refer: DOME by Bhandari, Fig. 15.11,case 2(a)

40. iii) Expected bearing life in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, countershaft of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) iv) L10 = 60𝑛L10h 106 = 60×138×25000 106 = 202.5 millions revolution since, intermediate shaft speed is running at 138 RPM v) Trial 1 Tentatively, we select Bearing 32207A. (Refer: PSG design DATA, pg.no. 4.25) vi) Bearing A: FrA = 1156.152N and FaA = 926.8N FaA FrA = 926.8 1156.152 = 0.8 > e = 0.37 P =Equivalent bearing load = X.Fr + Y.Fa = 0.4x1156.152+ 1.6x926.8 = 1945.34N C =dynamic capacity = P(𝐿10)0.3 x (Load factor) = 1945.34 (202.5)0.3 (1.5) = 14355.37N Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25)

41. vii) Similarly for Bearing B: FrB = 2378.689N and FaB = 743.34N Fa𝐵 Fr𝐵 = 743.34 2378.689 = 0.31 < e = 0.37 P = Equivalent bearing load = X.Fr + Y.Fa = 1x1156.152+ 0x926.8 = 1156.152N C = dynamic capacity = P(𝑳𝟏𝟎)𝟎.𝟑 x (Load factor) = 1156.152 (202.5)0.3 (1.5) = 8531.66N Hence, Bearing 32207A (C = 57600N) is suitable. (Refer: PSG design DATA, pg.no. 4.4 & 4.25) vii) The dimensions of bearing

42. • Selection of Keys for gear mounting on intermediate shaft i) Part No. is 22 and Part Name is Intermediate shaft key 1. ii) Function: Mounting bevel gear (G2) on intermediate shaft iii) TWO orthographic views of key iv) FBD of key

43. v) The martial selected for Key is plain carbon steel with grade C50 ,Because- This steel is suitable for making keys, shaft, cylinder, and machined components requiring wear resistance (Refer: PSG design data book, Page No. 1.9 & 1.10) vii) Properties of C50 : 1. Sut = 660 N/mm2 2. Syt = 380 N/mm2 viii) Factor of safety chosen is 1.5 because, key failure is more economical than gear or shaft failure in any extreme conditions. ix) Allowable stresses: 1. 𝜎 = 380 1.5 = 253.33N/mm2 and 2. 𝜏 = 0.5×380 1.5 = 126.67N/mm2 x) Shaft diameter for gear mounting = 1.1d = 38.5mm ; The dimensions of parallel key obtained from table is bxh = 12x8 and keyway depth is 6mm (Refer: PSG design data book, Page No. 5.24) xi) The effective length is found using empirical relations available for key ∴ l = 1.5×38.5 = 58mm < 84mm bxhxl = 12x8x58

44. xii) Checking for induced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×77887 12×8×𝟓𝟖 = 28N/mm2 < 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 56N/mm2 < 253.33N/mm2 Hence, design is safe i) Part No. is 23 and Part Name is Intermediate shaft key 2. iv) Take l = 25mm and check for induced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×77887 12×8×𝟐𝟓 = 64.90N/mm2 < 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 129.81N/mm2 < 253.33N/mm2 iii) Orthographic views , FBD, Material for key , Fos are same as earlier case Also dia. for gear mounting is same. Hence bxh = 12x8 only thing changes is length of key which should be < 27mm ii) Function: Mounting spur gear (S1) on intermediate shaft v) Hence, our design is safe bxhxl = 12x8x25

45. • Selection of circlips (External type) i) Part No. is 26 and Part Name is circlip 1 ii) Function: Locking bevel gear (G2) in its position on intermediate shaft iii) Detailed views with design notations iv) Commonly used steel for circlip is C65 (same as used in springs and washers) (Refer: PSG design DATA, pg.No. 5.3) v) Properties of C65: 1. Sut = 750 N/mm2 2. Syt = 430 N/mm2 (Refer: PSG design DATA, pg.No. 1.9 & 1.10)

46. vi) In our case d1 = 1.1d = 38.5mm and for this circlip selected from light series (A) is designated as Circlip, Light A 38 (Refer: PSG design DATA, pg.no. 5.4) Notati on Meaning Value (mm) d1 Shaft dia. For gear mounting 38 d2 Groove diameter o shaft for circlip 36 d3 Unexpanded dia. Of circlip 35.2 d4 Expanded dia. Of circlip 50.6 d5 Hole dia. For expanding circlip 2 s Thickness of ring in axial direction 1.5 b Thickness of ring in radial direction 5.8 a Height of ears of ring for expansion 4.2 v) Design dimension getting from catalogue for above mentioned circlip are vii) As per catalogue, Maximum axial load carry by these circlip = 29100N >> 926.8N so we can use it without any worry

47. • Design of shafts 2. Output shaft ii) Design of based on ASME code i) Part no. is 6 and part name is output shaft iii) Function: Transmission of torque and power to conveyor roller iv) TWO orthographic views

49. vii) FBD & BMD of shaft in respective plane ∴ Maximum bending moment at S2 = Mb = 79258.0642 + 216238.72 = 230,306.35 Nmm Also, Mt = 60 ×𝑃 2𝜋𝑁 = 60 ×0.75×106 2𝜋×23 = 311390.11 Nmm Considering load factor = 1.2 Hence, Mt = 1.2 x 311390.11 = 373668.13 Nmm

50. viii) Thus, Mb = 230306.35 Nmm & Mt = 373668.13 Nmm ix) Take Kb = Kt =1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 230306.35)2+(1.5 × 376668.13)2 ∴ 𝑑 = 35.61𝑚𝑚 But we haven’t considered the weight of gears while calculating diameter of shaft, hence we consider the shaft of diameter d = 44 mm xi) Finally, Shaft dia. for bearing mounting 0.8d & 1.1d 35.2mm & 48.4mm Shaft dia. for gear mounting d 44mm Shaft dia. for resting gear on one side 1.2d 52.8mm

51. xii) Checking stress concentration at critical sections To reduce the stress concentration at critical sections (X-X’, Y-Y’ and Z-Z’) suitable fillet radius of (0.09d) is provided. Where d is the diameter of shaft We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at these sections and then comparing with Endurance limit (Se ) If σ > Se ,then we have to change the dependent dimension such as 1.1d,0.8d,1.2d and if σ < Se , then we can say that Design is safe and correct. 0.09d 0.09d 0.09d

52. A) At section X-X’ for Bending and Torsion

53. B) At section Y-Y’ for Bending and Torsion

54. C) At section Z-Z’ for Bending and Torsion d For steel 40C8 Se’ = 0.5xSut = 290MPa Se = Ka.Kb.Kc.Kd.Se’ Ka = 0.84 (machined & cold-drawn) Kb = 0.85 (0.75<d<75mm) Kc = 1 (100% reliability) q = 0.95 (Notch radius = 4mm) 𝑟 𝑑 = 0.09𝑑 0.8𝑑 = 0.11 & 𝐷 𝑑 = 𝑑 0.8𝑑 = 1.25 Thus Kt for bending case Kt = 1.65 Refer: DOME by bhandhari, fig. 5.5) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.62 Kd = 𝐾𝑓 −1 = 0.62 Se = 0.84x0.85x1x0.62x290 Se = 128Mpa Bending stress at Z-Z’ is 𝝈𝒃 = 𝟑𝟐 𝑴𝒃 𝒛 𝝅 (𝟎. 𝟖𝒅)𝟑 𝑀𝑏 𝑧 = 31987 Nmm 𝜎𝑏 = 32×31987 𝜋 (35.2)3 = 7.5MPa < Se From BMD Similarly Kt for torsional case Kt = 1.4 (Refer: DOME by bhandhari, fig. 5.6) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.38 Kd = 𝐾𝑓 −1 = 0.72 Torsional shear stress at Z-Z’ is 𝝉 = 𝟏𝟔 𝑴𝒕 𝝅 (𝟎. 𝟖𝒅)𝟑 Se = 0.84x0.85x1x0.72x290 Se = 150Mpa 𝜏 = 16×373668.18 𝜋 (35.2)3 = 43.63MPa 𝜏 < Se i.e. Design is safe at section Z-Z’

55. • Selection of bearings 2. Output shaft bearings i) Part No. 9 and part name is Output shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1) iii) Radial load acting on bearings; FrA = 440.322 + 1201.3262 = 1279.48N FrB = 609.672 + 1663.3472 = 1771.56N FrB > FrA and Ka = 0 It matches with load case 2(a): from fig 15.22 of DOME by bhandhari

56. And here also, we are going with face-to-face configuration Hence axial load acting on bearings are given as FaB = 0.5 𝐹𝑟𝐵 𝑌 = 0.5 ×1771.56 1.6 = 553.61N and as Ka = 0 FaA = FaB iv) Expected bearing life in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, shafts of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) v) L10 = 60𝑛L10h 106 = 60×23×25000 106 = 34.5 millions revolution since, output shaft speed is running at 23 RPM

57. vi) Diameter for bearing mounting is 0.8d = 35.2mm (Non-gear side) Trial 1 For Bearing A : 32207A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 1397.56 x (34.5)0.3 x (1.5) = 6064.67N Hence, Bearing 32207A (C = 57600N) is suitable. 𝑭𝒂𝑨 𝑭𝒓𝑨 = 𝟓𝟓𝟑. 𝟔𝟏 𝟏𝟐𝟕𝟗. 𝟒𝟖 = 𝟎. 𝟒𝟑𝟐𝟔 > 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25)

58. vii) Diameter for bearing mounting is 1.1d = 1.1x44 = 48.4mm (Gear side) Trial 1 For Bearing B : 32210A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 1771.56 x (34.5)0.3 x (1.5) = 7687.63N Hence, Bearing 32210A (C = 72600N) is suitable. 𝑭𝒂𝑩 𝑭𝒓𝑩 = 𝟓𝟓𝟑. 𝟔𝟏 𝟏𝟕𝟕𝟏. 𝟓𝟔 = 𝟎. 𝟑𝟏𝟐𝟒𝟗 < 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) FrB = 1771.56N and FaB = 553.61N PB =Equivalent bearing load = X.FrB + Y.FaB = 1x1771.56+ 0x553.61 = 1771.56N

59. i) Part No. is 21 and Part Name is output shaft key ii) Function: Mounting spur gear (S2) on output shaft • Selection of Keys for gear mounting on Output shaft iii) TWO orthographic views iv) FBD of key

60. v) On output shaft torque acting = 311390.106Nmm which is very high, So here we go with higher grade of steel which is C60 (Refer: PSG design data book, Page No. 1.9 & 1.10) This steel is suitable for making hardened screws and nuts, couplings , machined spindles vii) Properties of C60 : 1. Sut = 750 N/mm2 2. Syt = 420 N/mm2 viii) Factor of safety chosen is 1.2 because, key failure is more economical than gear or shaft failure in any extreme conditions. ix) Allowable stresses: 1. 𝜎 = 420 1.2 = 350N/mm2 and 2. 𝜏 = 0.5×380 1.2 = 175N/mm2 x) Diameter of shaft for gear mounting = d = 44mm ; for these The dimensions of parallel key obtained from table is bxh = 14x9 and keyway depth is 6.5mm (Refer: PSG design data book, Page No. 5.24)

61. xi) Length of key is taken to be 30mm = width of S2 xii) Checking for allowable stresses: 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×311390.11 14×9×30 = 164.75N/mm2 < 175N/mm2 Also, 𝜎 = 2 𝜏 = 329.51N/mm2 < 350N/mm2 Hence, design is safe bxhxl = 14x9x30 i) Part No. is 27 and Part Name is circlip 2 ii) Function: Locking spurl gear (S2) in its position on output shaft iii) The diameter for S2 mounting is d = 44mm thus circlip selected from light series A is • Selection of circlips (External type) Circlip, Light A 45 Notation Value (mm) Notation Value (mm) d1 45 D5 2 d2 42.5 s 1.75 d3 41.5 b 6.7 d4 59.4 a 4.7 (Refer: PSG Design DATA, pg.no. 5.3 & 5.4)

62. • Design of shafts 3. Input shaft i) Part no. is 5 and part name is input shaft ii) Design of based on ASME code iii) Function: Transmission of torque and power from motor to G1

64. ∴ Maximum bending moment at bearing B = Mb = 29556.152 + 9447.752 = 31029.44Nmm Also, Mt = 60 ×𝑃 2𝜋𝑁 = 60 ×0.75×106 2𝜋×𝟔𝟗𝟎 =10379.67Nmm Considering load factor = 1.3 Hence, Mt = 1.3 x 10379.67 = 13793.5 Nmm

65. viii) Thus, Mb = 31029.44 Nmm & Mt = 12455.6 Nmm ix) Take Kb = Kt = 1.5 (Refer: Design book by Bhandari, Table 9.2, pg.no.334) x) By maximum shear stress theory, τ𝑚𝑎𝑥 = 16 ߨ𝑑3 (𝑘𝑏 . 𝑀𝑏)2+(𝑘𝑡 . 𝑀𝑡)2 = 16 ߨ𝑑3 (1.5 × 31029.44)2+(1.5 × 13493.5)2 ∴ 𝑑 = 15.1mm But we haven’t considered the weights of gears and bearings while calculating diameter of shaft, Also minimum dia. for which taper roller bearing is available is 30mm. hence, we consider the shaft of diameter d = 30mm xi) Finally, Shaft dia. for bearing mounting d 30mm Shaft dia. For coupling 0.8d 24mm Shaft dia. for gear mounting d 30mm Shaft dia. for resting gear on one side 1.2d 36mm

66. xii) Checking stress concentration at critical sections 1. From BMD it is clear that ; X-X’ is a critical section as bending moment is high over there 2. To reduce the stress concentration at critical sections (X-X’) suitable fillet radius of (0.08d) is provided. Where d is the diameter of shaft 3. We have to find bending (𝛔𝐛) and torsional (𝝉) stresses at these sections and then comparing with Endurance limit (Se ) 4. If σ > Se ,then we have to change the dependent dimension such as 0.8d,1.2d or 0.08d and if σ < Se , then we can say that Design is safe and correct.

67. A) At section X-X’ for Bending and Torsion For steel 40C8 Se’ = 0.5xSut = 290MPa Se = Ka.Kb.Kc.Kd.Se’ Ka = 0.84 (machined & cold-drawn) Kb = 0.85 (0.75<d<75mm) Kc = 1 (100% reliability) q = 0.84 (Notch radius = 2.5) Refer: DOME by Bhandari, fig. 5.23) 𝑟 𝑑 = 0.08𝑑 𝑑 = 0.08 & 𝐷 𝑑 = 1.2𝑑 𝑑 = 1.2 Thus Kt for bending case Kt = 1.8 Refer: DOME by Bhandari, fig. 5.5) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.672 Kd = 𝐾𝑓 −1 = 0.598 Se = 0.84x0.85x1x0.598x290 Se = 123.84Mpa Bending stress at X-X’ is 𝝈𝒃 = 𝟑𝟐 𝑴𝒃 𝒙 𝝅 (𝒅)𝟑 𝑀𝑏 𝑧 = 26466.32 Nmm 𝜎𝑏 = 32×26466.32 𝜋 (30)3 = 9.98MPa < Se From BMD Similarly Kt for torsional case Kt = 1.45 (Refer: DOME by Bhandari, fig. 5.6) Kf = 1 + 𝑞 𝐾𝑡 − 1 = 1.378 Kd = 𝐾𝑓 −1 = 0.7257 Torsional shear stress at Z-Z’ is 𝝉 = 𝟏𝟔 𝑴𝒕 𝝅 (𝒅)𝟑 Se = 0.84x0.85x1x0.7257x290 Se = 150.26Mpa 𝜏 = 16×13493.5 𝜋 (35.2)3 = 2.545MPa 𝜏 << Se i.e. Design is safe at section X-X’

68. • Selection of bearings 3. Input shaft bearings i) Part No. 8 and part name is Intput shaft bearing ii) Selected type of bearing for given application is Taper roller bearing, because it has following advantages- It can carry combine radial and axial loads (Refer: PSG Data book, pg.no. 4.1) iii) Radial load acting on bearings; FrA = 347.722 + 111.152 = 365.052N FrB = 861.742 + 294.62 = 910.71N FrB > FrA and Ka = 36.7N It matches with load case 2(a): from fig 15.22 of DOME by Bhandhari

69. And here also, we are going with face-to-face configuration Hence axial load acting on bearings are given as FaB = 0.5 𝐹𝑟𝐵 𝑌 = 0.5 ×910.71 1.6 = 284.6N and as Ka = 36.7N FaA = FaB + Ka FaA = 321.3N iv) Expected bearing life in hrs. L10h = 25000hrs Machines for use 8hrs/day ad fully utilized – E.g. cranes for bulk goods, ventilating fans, shafts of gearboxes, the recommended life is from 20000 to 30000h (Refer: PSG design DATA, Table-Life of bearing, pg.no. 4.5) v) L10 = 60𝑛L10h 106 = 60×690×25000 106 = = 1035 millions revolution Since, Input shaft speed is running at 690 RPM

70. vi) Diameter for bearing mounting is d = 30mm (Non-Gear side) Trial 1 For Bearing A : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PAx(𝐿10)0.3 x (Load factor) = 660.1 x (1035)0.3 x (1.5) = 7946.64N Hence, Bearing 32206A (C = 43800N) is suitable. 𝑭𝒂𝑨 𝑭𝒓𝑨 = 𝟑𝟐𝟏. 𝟑 𝟑𝟔𝟓. 𝟎𝟓𝟐 = 𝟎. 𝟖𝟖 > 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) PA = Equivalent bearing load = X.FrA + Y.FaA = 0.4x365.052 + 1.6x321.3 = 660.1N FrA = 365.052N & FaA = 321.3N

71. vii) Diameter for bearing mounting is d = 30 mm (Gear side) Trial 1 For Bearing B : 32206A is selected (Refer: PSG design DATA, pg.no. 4.25) C = dynamic capacity = PBx(𝐿10)0.3x (Load factor) = 910.7 x (1035)0.3 x (1.5) = 10963.5N Hence, Bearing 32206A (C = 46800N) is suitable. 𝑭𝒂𝑩 𝑭𝒓𝑩 = 𝟐𝟖𝟒. 𝟔 𝟗𝟏𝟎. 𝟕𝟏 = 𝟎. 𝟑𝟏𝟐𝟓 < 𝒆 = 𝟎. 𝟑𝟕 (Refer: PSG design DATA, pg.no. 4.4 & 4.25) FrB = 910.71N and FaB = 284.6N PB = Equivalent bearing load = X.FrB + Y.FaB = 1x1771.56+ 0x553.61 = 910.71N

72. • Locking arrangement for bearing A (of input shaft only) Lock washer Ring Nut iv) We have to find suitable Lock washer and Ring nut from manufactures catalogue which totally depends on shaft dia. At these section i) Part No. are 29,30 and Part name is Lock washer and Ring nut respectively. ii) Function: Locking bearing A from other end to restricts it’s movement in axial direction iii) The shaft diameter for lock washer and ring nut is 0.8d = 0.8x30 = 24mm

73. v) Finally, the standard elements select from PSG design data tables Notations Values (mm) h 23 d1 32 d2 42 a 5 b 5 t 1.25 N 13 (Refer: PSG design DATA, pg. No. 5.86) Notations Values (mm) d 25 D 38 b 7 s 4 t 2 M25 x 1.5 (Refer: PSG design DATA, pg. No. 5.85)

74. i) Part No. is 20 and Part Name is Input shaft key v) Check for induced stresses; 𝜏 = 2𝑇 𝑏ℎ𝑙 = 2×13493.5 8×7×𝟑𝟎 = 16.72N/mm2 << 126.67N/mm2 Also, 𝜎 = 2 𝜏 = 33.44N/mm2 << 253.33N/mm2 iv) Orthographic views , FBD, fos = 1.5 are same as earlier case Diameter for gear mounting is d = 30mm, Hence bxh = 8x7 and keyway depth is 1.7mm also, length of key = 𝑙 = 30 mm < 60mm ii) Function: Mounting Bevel gear (G1) on input shaft vi) Hence, our design is safe bxhxl = 8x7x45 • Selection of Keys for gear mounting on Input shaft iii) The martial selected for Key is plain carbon steel with grade C50 1. Sut = 660 N/mm2 2. Syt = 380 N/mm2

75. i) Part No. is 28 and Part Name is circlip 3 ii) Function: Locking bevel gear (G1) as well as bearing B iii) The diameter for G1 mounting is d = 30mm thus circlip selected from light series A is • Selection of circlip (External type) Circlip, Light A 30 (Refer: PSG Design DATA, pg.no. 5.3 & 5.4) Notation Value (mm) Notation Value (mm) d1 30 d5 2 d2 28.6 s 1.5 d3 27.9 b 3.5 d4 41 a 5

76. • Locking arrangement for G1 from open end iv) Lock washer and Ring nut assembly is same as that of selected on slides 72 & 73 because mounting dia. is same in both the cases i.e 24mm i) Part No. is 31 and Part name is Washer ii) Function: Evenly distributing load from nut to larger area iii) The shaft dia. for washer is 24mm, thus dimension obtained from manufactures catalogue are Washer Lock washer & Ring nut assembly Notation Values (mm) d 24 D 48 s 7

77. i) Part No. is 17 and Part Name is Input shaft oil seal ii) Function: Prevents oil leakage from input end of gearbox iii) Diameter for oil seal mounting is 0.8d = 0.8x30 = 24 mm iv) Also Type A oil is selected for input shaft since torque acting on it is quite low as well as it running at high speed Type A : Stiffener (metal) ring is covered with rubber elastomer to tolerate thermal expansion • Selection of Oil seal for Input shaft v) Thus the oil seal selected for 24mm shaft dia. from catalogue have dimension of Notation Value b 7 c (min) 0.3 Nominal housing diameter 35 (Refer: PSG Design DATA, pg. No. 5.11) Note: Find fit allowance and tolerances for Type A seals , see table on pg. No. 5.112

79. i) Part No. is 18 and Part Name is Output shaft oil seal ii) Function: Prevents oil leakage from input end of gearbox iii) Diameter of shaft R for oil seal mounting is 1.1d = 0.8x44 = 48.4 mm iv) Also Type B oil is selected for output shaft since torque acting on it is very high compared to input shaft Type B : Stiffener (metal) ring is at outer periphery enclosing elastomer material • Selection of Oil seal for Output shaft v) Thus the oil seal selected for 48.4mm shaft dia. from catalogue have dimension of Notation Value b 8 c (min) 0.4 Nominal housing diameter 62 (Refer: PSG Design DATA, pg. No. 5.11) Note: Find fit allowance and tolerances for Type B seals , see table on pg. No. 5.112

80. • Design of Casing 1. Lower casing  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions

81. • Design of Casing 1. Upper casing  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions

82. • Design of bearing caps  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions

83. • Design of stuffing box  FBD  2 orthographic views with design notation  Material  Factor of safety  Calculations  Table of final dimensions

84. • Weight of gearbox Total weight of gearbox = 120 + 3.6 + 5.4 + 10 + 34.4 + 12 = 𝟏𝟖𝟔𝒌𝒈

85. • Selection of Eye Bolts i) Part No. is 28 and Part Name is eye bolt ii) Function: Lifting of gearbox using cranes/hoists iii) Orthographic Views of eye bolt iv) Load acting on eye bolt = Overall weight of gearbox when lifted (Pw) = 186 x 10 = 1860N iv) The material selected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts,bolts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) 1. Sut = 580/mm2 2. Syt = 300N/mm2

86. v) Factor of safety is taken to be 7, since as gearbox get lifted all the weight is come on eye bolt Allowable stress calculation: σ = 300 7 = 42.85MPa vii) Critical mode of failure : Shearing of threads of eye bolt σ = 4𝑃𝑤 𝜋𝑑𝑐 2 𝑑𝑐 2 = 4𝑃𝑤 𝜋σ ∴ 𝑑𝑐 2 = 4 × 1860 𝜋 × 42.85 𝑑𝑐 = 7.43𝑚𝑚 𝑑 = 𝑑𝑐 0.86 = 8.64mm M10 x 0.75 is selected. Hence,

87. • Selection of Nut-Bolt for casing i) Part No. is 33,34 and Part Name is casing bolt & casing nut (Hexagonal Nut and Bolt) ii) Function: Joining of upper and lower casings iii) Critical mode of failure : Tensile iv) Material : 30C8 and fos = 6 M8 x 0.5 is selected. (Refer: PSG design DATA, pg.no.5.49 & 5.50 ) Such 6 Nut-bolt pair is needed

88. The same size of screw(below figure) are used nearby bearing region M8 x 0.5 Screw is selected. For 4 bearing semi-circular regions we requiring 8 screw (Refer: PSG design DATA, pg.no.5.61 & 5.62 ) Again, 6 same hexagonal socket head cap screw [M8 x 0.5] is used for attachment of stuffing box to the casing The nut-bolts, screws used to join two casing are intentionally selected of fine series to ensure proper tightening and thus avoiding oil leakage

89. • Selection of foundation Bolts i) Part No. is 32 and Part Name is Foundation bolt ii) Function: Fixing/Mounting of gearbox on support iii) Load acting on foundation bolt = Overall weight of gearbox when lifted (Pw) = 186 x 10 = 1860N iv) Critical mode of failure : Compression and shearing of threads v) The material selected for shaft is Plain carbon steel with grade 40C8 because, this steel is used for making shafts, bolts, crankshaft, automobile axle (Refer: PSG design data book, pg.no. 1.9, 1.10) 1. Sut = 580/mm2 2. Syt = 300N/mm2

90. v) Factor of safety is taken to be 6, since as gearbox get lifted all the weight is come on eye bolt Allowable shear stress calculation: τ = 0.5×300 5 = 30MPa 𝑑𝑐 2 = 4𝑃𝑤 𝜋𝜏 𝑑𝑐 = 8.88𝑚𝑚 𝑑 = 𝑑𝑐 0.86 = 10.32mm M12 x 1.25 The next standard bolt is selected of, 𝜏 = 4𝑃𝑤 𝜋𝑑𝑐 2 ∴ 𝑑𝑐 2 = 4 × 1860 𝜋 × 30 On dimensional constraint basis the screw selected for- 1. bearing cap attachment to housing and 2. cover plate attachment to stuffing box is of M5 x 0.5 6 for each cap x 5 = 30 Hexagonal socket head cap screw are used.

Gearbox design

Gearbox design

Recommandé

Contenu connexe

Similaire à Gearbox design

Similaire à Gearbox design(20)

Dernier

Dernier(20)

Gearbox design